The NIST Handbook of Mathematical functions eq. 10.19.1 [ref] gives for the asymptotic behavior of the Bessel function $J_\nu(z)$ for large real order $\nu\to\infty$ for fixed argument $z$ as
$$
J_\nu(z) \sim \frac{1}{\sqrt{2\pi \nu}}\big(\frac{ez}{2\nu}\big)^\nu
$$
without providing any error. Is it known what the corrections are to this asymptotic relation?
Best Answer
We can take the integral representation $$J_\nu(z) = \frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}} \int_{-1}^1 (1 - t^2)^{\nu - 1/2} \cos z t \, dt,$$ which is valid for $\nu > -1/2$. The change of variables $-u^2 = \ln( 1- t^2)$ gives $$f(u) = \frac {\left| u \right| e^{-u^2/2} \cos z \sqrt {1 - e^{-u^2}}} {\sqrt {1 - e^{-u^2}}}, \\ J_\nu(z) = \frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}} \int_\mathbb R f(u) e^{-\nu u^2} du.$$ Expanding the gamma function around $\nu = \infty$ and expanding $f(u)$ around $u = 0$, we obtain as many terms in the asymptotic series as we want: $$\frac {z^\nu} {2^\nu \hspace {1px} \Gamma {\left( \nu + \frac 1 2 \right)}} = \frac 1 {\sqrt {2 \pi}} \left( \frac {e z} {2 \nu} \right)^\nu \left( 1 + \frac 1 {24} \nu^{-1} + O(\nu^{-2}) \right), \\ f(u) = 1 - \frac {2 z^2 + 1} 4 u^2 + O(u^4), \\ \frac 1 {\sqrt \pi} \int_{\mathbb R} f(u) e^{-\nu u^2} du = \nu^{-1/2} - \frac {2 z^2 + 1} 8 \nu^{-3/2} + O(\nu^{-5/2}),\\ J_\nu(z) = \frac 1 {\sqrt {2 \pi}} \left( \frac {e z} {2 \nu} \right)^\nu \left( \nu^{-1/2} - \frac {3 z^2 + 1} {12} \nu^{-3/2} + O(\nu^{-5/2}) \right).$$